Date: 01/17/97 at 16:59:36 From: John Hartge Subject: Slope-intercept forms My daughter is having trouble with the slope intercept form of lines. Given a line in this form, she has trouble finding a line perpendicuar or parallel to it. She also doesn't understand how to find the x and y intercepts from an equation like 4x + 3y = 6 or its graph. When she was doing mixture problems earlier this year, she kept having trouble writing a table to figure it out. Can you please help me? Thanks, John
Date: 01/21/97 at 22:46:33 From: Doctor Keith Subject: Re: Slope-intercept forms Hi, That is a pretty good list of things; glad to see a dad helping his kid! Hopefully I will be able to clear this up. I have written it so your daughter can follow it (it is a bit long since I am quite wordy, so you probably will want to print it out). I have tried to guess at exactly what she knows and does not know, so please reply if something needs clarification and forgive things that might be too simple. Now we are lucky in that I think we can handle most of the questions by doing a good review of equations of a line. The slope intercept form of a line is: y = mx + b y is the vertical measurement (sometimes called dependant variable) x is the horizontal measurement (sometimes called independant variable) m is the slope (the change in the y divided by the change in x which is sometimes expressed as rise over run) b is the y-intercept I advise you to put any linear equation into this format so that you can then use some standard rules and tools to get your answers. Thus if you had a problem like: 3y = 5/3x - 8 you would rewrite it to read (by dividing by 3): y = 5/9x - 8/3 Now let's tackle the easiest part of the problem, which is finding the x and y intercepts. My math teacher taught us what he called the Watergate method (for obvious reasons as you will see). The method goes like this: To find the y-intercept: 1) Take your hand (or finger or whatever) and cover-up the x term (that is mx). (The cover-up is the reason for the name Watergate.) 2) The solution is y = b (x = 0). Note this works because with x = 0, we have mx = 0 and thus y = b, and since the y intercept is the value of y when x = 0, we are done. To find the x-intercept: 1) Take your hand (or finger or whatever) and cover-up the y term (that is just y this time). 2) We have 0 = mx + b. 3) Thus solving for x we have x = -b/m. Note this works because with y = 0, we have step (2), and the x-intercept is the value of x when y = 0 so again we are done. Thus in the problem above we have: y = 5/9x - 8/3 y-intercept = b = -8/3 x-intercept = -b/m = (8/3)(9/5) = 24/5 Now the next easiest thing is getting a parallel line. Actually, since we know how to get the intercepts, we can get the parallel lines just as easily. Note that all parallel lines have the same slope. This makes sense because both lines must increase the same amount in y for a given change in x or at some point they would meet (this is easy to convince yourself of by drawing some lines on a piece paper). Now if you draw a bunch of parallel lines, you will notice the only difference between them is where they intercept the x or y axis. So by changing, say, the y-intercept, you can create a new parallel line. We know that the y-intercept is b in our equation so to get a parallel line you just need to change b. Example of line parallel to y = 5/9x - 8/3: y = 5/9x - 1 Now say you wanted a line parallel through a point (a bit trickier but still no problem). You are given (x1,y1) and you need to find a b1 for which y1 = mx1 + b1 is valid. Thus you can solve for b1 and you get: b1 = y1 - m(x1) Example line parallel to y = 5/9x - 8/3 through (9,1): b = 1 -5/9(9) = -4 y = 5/9x -4 Okay, now you need to know how to find perpendicular lines also. Even more fun. This relies on the neat property of slopes that the product of the slopes of two perpendicular lines is -1. The easiest way I know to prove this involves some higher math that you don't want to deal with at the moment (no reason to muddy the waters) but you can see that this makes sense by the following: 1) Draw two perpendicular lines on a blank sheet of paper (no axis please - we are going to rotate the paper to compare slopes). 2) We shall agree that all lines are drawn left to right regardless of the up/down motion. 3) Thus we have that lines are drawn UP if they start in the lower left and go to the upper right. Note these lines will have positive slopes (the y value increases as x increases). 4) Similarly, lines that are drawn DOWN (upper left to lower right) have negative slopes (the y values decrease if x increases). 5) Note that a negative times a positive must be a negative, so we know from (3) and (4) that the products of the slopes must be negative. 6) Rotate the paper in front of you. You will notice that as one line gets steeper (large change in y for a given change in x), the other line gets shallower (small change in y for a given change in x). This indicates that the slopes are inversely proportional. In other words, the product of the slopes of the lines should be a constant regardless of how you rotate the paper. 7) So far, we have that the product of the slopes should be a negativeconstant. To see why the constant is -1, do the following. Turn the paper so that the lines are at a 45 degree angle with the horizontal (it should look like an x). Now the slopes are 1 and -1 for this case. We have noted that the slopes are a constant regardless of the rotation [i.e., the product of the slopes gives the same answer for our 45 degree case as for any other rotation angle (except for a constant line y = 0 and x = 0, since x = 0 is not allowed in math)]. Since the product in this case is -1, it must hold for the other cases. As I indicated earlier, while this does not prove it (the proof is beyond the scope of this discussion) it helps you to see that it is true. Now, using this you can get a perpendicular line by the following: let your first line be y = m1x + b1 you want a line y = m2x + b2 that is perpendicular Solve by using the fact that the product of the slopes is -1: m1(m2) = -1 m2 = -1/m1 Then if you want the line to go through a particular point (x1,y1), you can find b2 by (similar to the parallel case): y1 = m2x1 + b2 b2 = y1 - m2x1 = y1 + x1/m1 If you do not have to go through a particular point, pick any number for b2 (I suggest b2 = 0). As for general graphing and writing the tables to graph form, I suggest the following: 1) Pick equally spaced numbers usually centered on the origin to use as your x values. Try to pick them so your calculations are easy (i.e., for a slope of 1/3 values of -3,0,3,6,... would be great). 2) Make a table that looks like x | y and put the x values -------------- in the left column | | | 3) For each value in the x column, calculate y = mx + b for y (i.e., you know m, x, and b) and put this in the y column (this can be simplified in linear equations by noting that by multiplying the change in x times the slope and adding it to the previous value you will get the new value, see example below). 4) When you are done with the table, just plot the points and connect the dots. Example: y = 2/3 x - 4 1) I pick -3,0,3,6,9 2) x | y ---------------- -3 | 0 | 3 | 6 | 9 | 3) x | y ---------------- -3 | 2/3(-3) -4 = -6 0 | 2/3(0) - 4 = -4 (or -6 + 2/3(0--3) = -6 +2 ) 3 | 2/3(3) - 4 = -2 (or -4 + 2/3(3-0) = -4 +2 ) 6 | 2/3(6) - 4 = 0 (or -2 + 2 ) 9 | 2/3(9) - 4 = 2 (or 0 + 2 ) So you have: x | y ---------------- -3 | -6 0 | -4 3 | -2 6 | 0 9 | 2 Hope this helps. Keep up the good work and let me know how it all works out. We are here to help. Good luck! -Doctor Keith, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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